# I Definition of discrete Subgroup quick q

1. Jan 19, 2017

### binbagsss

Hello,

Just a really quick question on definition of discrete subgroup.
This is for an elliptic functions course, I have not done any courses on topology nor is it needed, and most of the stuff I can see online refer to topology alot, so I thought I'd ask here.

I need it in the complex plane

I have : A subset $S$ of a topological space is called discrete if for any $s \in S$ there is an open set $U$ s.t

$U \cap S = \{s\}$

So can someone clarfiy this, in really simple terms please, I'm not too familiar with such notation ...

Is $s$ a single element?
So it is saying that the intersection of $U$ and $S$ is the set of elements ${s}$, or just the element $s$, I'm unsure what the curled brackets {} denote.

Many thanks

2. Jan 19, 2017

### Orodruin

Staff Emeritus
The curly brackets enclose the elements of a set, in this case the set $U \cap S$. The element $s$ is a single element (but you should be able to select it to be any element in $S$). The statement just says that the subset is discrete if there for every $s$ in $S$ exists an open set in the topological space such that $s$ is the only element in $S$ that belongs to it.

3. Jan 19, 2017

### Staff: Mentor

The brackets denote a set. $U$ and $S$ are sets, so is $U \cap S$. In case $S$ is discrete, $U_s \cap S = \{s\}$ the set which contains exactly one element $s$. I wrote $U=U_s$ because different $s$ result in different open neighborhoods $U_s$.
An example for a discrete set in the complex plane is $S=\mathbb{Z} +i\mathbb{Z}$, the knots of a lattice. Around a knot $s$, there is a neighborhood $U_s$ without any other points of $S$.

4. Jan 21, 2017

### Stephen Tashi

To emphasize the importance of how "open set" restricts the choice of $U$, consider the vertical line $x = 1$ in the xy-plane. The set $S$ of points on that line is not discrete. Let $U$ be the set of points on the perimeter of the unit circle. Then $U \cap S = {(1,0)}$, which is a single point. However, the set of points on the perimeter of the unit circle is not an open set. A crude intuitive notion of an open set in the xy plane is that it is "the insides" of one or more closed geometric figures , but not including their perimeters. For example, the set of points within the unit circle is an open set. The set of points within the unit circle plus some points on the perimeter of that circle is not an open set.

With that intuition of an open set, you can appreciate the intuitive idea that if an open set "hits" the line S then it must hit it at more than one point. For example, for any $\epsilon > 0$ the interior of any circle with center $(0,0)$ and radius $r > 1 + \epsilon$ contains more than one point of the line $S: \{(x,y): x = 1\}$.

If we now let $S$ be the discrete set of points $\{(1,0) ,(1,1),(1,2)\}$ then for each point in $S$ we can find some circle whose interior contains only that point. (Of course, we are free to pick circles with centers different that (0,0).)

5. Jan 21, 2017

### Orodruin

Staff Emeritus
Just to say that all of this depends on the topology imposed on the original set. A topological space is not only a set, it also contains a definition of what is considered to be an open set. What Stephen just said is true for the usual topology on $\mathbb R^n$. However, there are other choices, such as the discrete topology where any subset of the full set is an open set. In the discrete topology, any subset of the topological space would therefore be a discrete subset.

6. Jan 21, 2017

### binbagsss

Okay thanks, to just to confirm my understanding, if I have understood correctly, for every point on a circle, $s$ due to it's curvature you can find a set $U$ such tthat $U \cap S = {s}$, but this doesn't 'count' as it's curvature also means that no other points are contained within $U$ ?

7. Jan 21, 2017

### Orodruin

Staff Emeritus
No, it does not work because $U$ is not an open set. The point was that for any point on the line, you cannot find an open set that contains that point but does not contain any other points on the line.

8. Jan 21, 2017

### Staff: Mentor

It doesn't have anything to do with curvature. It is just an example which happens to be curved. For any point you can always find a set $U$ such that $U \cap S = \{s\}$. Only take $U=\{s\}$. This doesn't mean anything.
First of all, $U$ has to be open.
This rules out my example, except in the case of a discrete topology (see post #5).
Next, the point $s$ has to be alone, i.e. no other point of $S$ lies in $U$. It is nothing said about the points in $U$ other than $s$ is one of them. Sometimes it's the only one as in the discrete topology, and sometimes it's an entire inner area of a circle as with the metric induced topology of the Euclidean plane.
The shape of $U$ isn't important either.
It's only that open circles (as in post #4) often form a basis of the topology (as in the usual topology of $\mathbb{R}^n$, see explanation in post #4 and #5). I gave you an example of a discrete set $S$ in the complex plane or if you like $\mathbb{R}^2$ where you can think about differently shaped $U$ (see post #3).

9. Jan 21, 2017

### binbagsss

So the example of the perimenter did not count because it was taking $U=\{s\}$ only?

10. Jan 21, 2017

### Staff: Mentor

I assume we are talking about the following situation:

The topological space is the Euclidean real plane and the set $S_1$ is the perimeter of a circle and $s$ a point of $S_1$.

Now every open set $U$ that contains $s$ has automatically another point $s' \in S_1$ in it. No matter how you shape $U$ you cannot manage to have only one perimeter point in it. The only way out would be $U=\{s\}$, but this is not an open set in the given topology.

If we are talking about the line $S_2=\{(x,y)\in \mathbb{R^2}\,\vert \,x=1\}$ as our set, then the situation is completely the same:

Now every open set $U$ that contains $s$ has automatically another point $s' \in S_2$ in it. No matter how you shape $U$ you cannot manage to have only one line point in it. The only way out would be $U=\{s\}$, but this is not an open set in the given topology.

This shows that neither $S_1$ nor $S_2$ are discrete (in the topology induced by the metric). Especially curvature doesn't play a role!

An example for a discrete set would be $S_3 = \{(x,y)\in \mathbb{R^2}\,\vert \, x,y \in \mathbb{Z}\}$, the knots of the integer lattice. Here we can define for each point $s = (x_s,y_s) \in S_3$ a open neighborhood, i.e. open set $U_s = \{(x,y) \in \mathbb{R^2} \,\vert \, (x-x_s)^2-(y-y_s)^2<1 \}$, which is the inner of a unit circle with radius $1$ and center $(x_s,y_s)$, that does not contain any other point $s'=(x_0,y_0)$ with integer coordinates, i.e. points of $S_3$. Thus $U_s \cap S_3 = \{s\}$ for any $s\in S_3$ and $S_3$ is discrete.

The fact that I have chosen the inner area of circles for $U_s$ is obviously only, because it is convenient: to define and to prove. You may decrease the radius, or shape it differently. There are plenty of possibilities to define an open set $U_s$ which doesn't contain more than a single point $s \in S_3$ for the lattice gives us enough space to do so. All knots of the lattice are kind of alone, i.e. discrete.

11. Jan 21, 2017

### Stephen Tashi

If you are using "curvature" to mean "perimeter" then yes. But what happens in not due to the fact that the perimeter of a circle has a curved shape. For example, if we let $U$ be the triangle with vertices (0,-1),(0,1),(1,0) then the perimeter of $U$ intersects the line $x = 1$ in a single point. But you can't find a triangle whose interior (sans the perimeter) intersects the line in a single point.